Question: Subtract. $\dfrac{8}{6} - \dfrac{1}{8} = $
Solution: Before we can subtract our fractions, they need to have the same denominator. $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\frac{1}{8}$ $\dfrac{8}{6}$ $\dfrac{1}{8}$ $\dfrac{8}{6}-\dfrac{1}{8}$ Let's look at the multiples of each denominator and see which multiples they have in common. Denominator Multiples ${6}$ $6, 12, 18, \underline{24}$ $8}$ $ 8, 16, \underline{24}$ The least common denominator is ${24}$. Let's use multiplication to make each fraction have a denominator of $24$. ${\dfrac{8}{6}}=\dfrac{{8} \times {4}}{{6} \times {4}} = {\dfrac{32}{24}}$ $\dfrac{1}{8}}=\dfrac{1} \times 3}{8} \times 3} = {\dfrac3}24}}$ Now, we can subtract ${\dfrac{32}{24}} - \dfrac{3}{24}}$. $\dfrac{32}{24}$ $\dfrac{3}{24}$ $\dfrac{32}{24} - \dfrac{3}{24}$ $=\dfrac{{32}-3}}{24}$ $= \dfrac{29}{24}$ ${\dfrac{8}{6}} - \dfrac{1}{8}} = \dfrac{29}{24}$ We can also write $\dfrac{29}{24}$ as $1\dfrac{5}{24}$.